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User blog:DontDrinkH20/Some ZFC-Definable very big LCOs - uncomputable?
This time I'm going to be defining some really big ZFC-definable countable ordinals (some even provably exist). These are based on the principle of making ordinals which are not definable within another very "wide" structure, like a transitive model of ZFC or \(V_\kappa\) for large enough \(\kappa\). Ordinals ZFC Prove to Exist Let \(\varphi\) be a formula and let \(x\) be a set. \(\varphi:x\) if and only if \(\varphi(A)\) if and only if \(A\in x\) for all \(A\). Then, \(\theta_\alpha\) is defined as the smallest ordinal \(\beta\) such that there is no formula \(\varphi\) such that \(\varphi^{V_\alpha}:\beta\). That is, the smallest ordinal which is not definable within \(V_\alpha\). Given a transitive structure \(\mathcal{M}=(M;\in)\), \(\theta(\mathcal{M})\) is the analogous ordinal. \(\theta_\omega\) clearly doesn't exist; every ordinal in \(V_\omega\) is a finite number definable using the definable successor function. However, \(\theta_{\omega_1}\) clearly does exist because it contains uncountably many ordinals while there are only countably many formulae; not each formula can define a unique ordinal so some ordinals share definitions. Because the definition only quantifies over first-order formula bounded in \(\omega_1\), ZFC can actually state (and indeed prove) the existence of this ordinal. \(\mathfrak{s}\) (or "The synthetic constant") is then defined as the smallest ordinal such that \(\theta_{\mathfrak{s}}\) exists; i.e. there is some undefinable ordinal in \(V_{\mathfrak{s}}\). It has been established that \(\omega<\mathfrak{s}\leq\omega_1\), but questions of the exact value, or even whether \(\mathfrak{s}<\omega_1\), is unknown. A similar value, \(\mathfrak{t}\) is defined as the smallest ordinal such that \(\theta(L_{\mathfrak{t}})\) exists; i.e. there is an ordinal indiscernible in \(L_{\mathfrak{t}}\). It is easy to show that if \(\alpha\) is an \(L\)-inaccessible ordinal, then it is at least as big as \(\mathfrak{t}\) (and therefore if \(0^{\#}\) exists then \(\mathfrak{t}\) is finite). Here's why: *Let \(\alpha\) be \(L\)-inaccessible. In \(L\), \(V_{\alpha+1}\models\text{MK}\). Because MK proves there are indiscernable ordinals in \(V\), there are indiscernables in \(V_\alpha\) in \(L\). Intrestingly, this is equivalent to \(V_\alpha^L\) having indiscernables in \(V\). *\(V_\alpha^L\) has indiscernables. However, in \(L\), because \(\alpha\) is inaccessible, \(V_\alpha=L_\alpha\). This is a problem because now \(V_\alpha^L=L_\alpha^L=L_\alpha\). So, we have the result that \(L_\alpha\) has indiscernables and therefore is at least as big as \(\mathfrak{t}\). What is known is that \(\mathfrak{s}\) must be a limit ordinal. This is simple to prove; if \(\alpha\) is an indiscernable in \(V_{\beta+1}\), then it is either also an indiscernable in \(V_{\beta}\) (and therefore \(\theta_\beta\) exists) or \(\alpha=\beta\). Clearly, \(\alpha\neq\beta\) because \(V_{\beta+1}\) defines \(\beta\) as the largest transitive set well-ordered by \(\in\). Furthermore, \(\mathfrak{s}\) must be a limit of limit ordinals. Assume \(\theta_\alpha\) does not exist. Then, every ordinal \(\beta\in\alpha\) has a definition \(\varphi_\beta\) bounded in \(V_\alpha\). It is simple to see that \(V_{\alpha+\omega}\) can define \(\alpha\) and therefore can define \(V_\alpha\), and therefore can define all \(\beta\in\alpha\) by bounding each \(\varphi_\beta\) to \(V_\alpha\). Therefore, \(\theta_{\alpha+\omega}\geq\alpha\). However, \(V_{\alpha+\omega}\) can easily define \(\alpha\) and therefore \(\alpha+\omega>\theta_{\alpha+\omega}>\alpha\); but that's impossible because then it would have to be a successor, even though it is a limit ordinal. It is also known that it cannot be eventually writeable and indeed \(\mathfrak{s}>\zeta\). This is because \(V_{\omega^2}\subset V_{\mathfrak{s}}\) can already define \(\mathcal{O}^{++}\) meaning we have a definition for every ordinal \(\alpha\) definable using \(\mathcal{O}^{++}\) on a natural number in \(V_{\mathfrak{s}}\). As a result, there must be some ordinal not definable using \(\mathcal{O}^{++}\) on any natural number (\(\zeta\) must be one of these ordinals) and as a result \(\mathfrak{s}>\zeta\). At this point I doubt that \(\mathfrak{s}\) is even countable. If you can somehow prove that it is even consistent to be countable, I will be amazed. Other Ordinals Let \(\Xi\) be the class of all ordinals \(\alpha\) such that there exists a transitive standard model \(\mathcal{M}\models\text{ZFC}\) with \(\omega_1\subset M\) such that \(\theta(\mathcal{M})=\alpha\). Then, \(\mathfrak{x}\) is the least ordinal in \(\Xi\). Of course, this assumes that there exists some transitive standard model of ZFC, and not only that but there is one which contains every \(\omega_1\). For this reason, ZFC doesn't even prove that this ordinal exists, although it can state it's existence in first-order logic because it is only undefinable on a set of formulae which are bounded in quantifiers. This can be seen as "the minimal indiscernable ordinal of a model which is right about the value of all countable ordinals." For example: if there exists a worldly cardinal \(\kappa\), \(\mathfrak{x}\) is bounded by \(\theta_\kappa\) (which is of course countable). This ordinal doesn't really have much relevance, but it is interesting nonetheless. 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